ACADEMIA SINICA
Volume 32, Number 3, September 2004
ON THE POSITIVE SOLUTIONS OF THE DIFFERENTIAL
EQUATION u′′− up = 0
BY
MENG-RONG LI(李明融)
Abstract. In this paper we work with the ordinary equa-tion u′′
−up= 0 and obtain some interesting phenomena concern-ing blow-up, blow-up rate, life-span, zeros, critical points and the asymptotic behavior at infinity of solutions to this equation.
Introduction. In our papers [1, 2, 3] we studied the semi-linear wave equation u + f (u) = 0 under some conditions, and we found some inter-esting results on blow-up, blow-up rate and the estimates for the life-span of solutions, but no information on the singular set. Here we want to deal with the particular cases in lower dimensional wave equations. We hope that the experiences gained here will allow us to deal with more general lower dimension later.
Consider stationary, one-dimensional semilinear wave equation
u′′ − up= 0, p ∈ (0, 1) , u (0) = 0 = u′ (0) .
After some computations one can find that the equation has infinite many
Received by the editors July 14, 2003. AMS Subject Classification: 34.
Key words and phrases: Estimate, life-span, blow-up, blow-up rate, asymptotic behav-ior.
This work is financed by NSC grant 87-2115-M-004-001.
solutions given by uc(t) = 0, t ∈ [0, c] , cp(t − c) 1 1−p, t > c,
where cp = (1 − p)2/(1−p)(2p + 2)1/(1−p). Thus, in particular, the solutions
of the above equation in general are not unique. It is clear that these func-tions up, p ≥ 1, u ≥ 0 are locally Lipschitz, and by the standard theory, the local existence and uniqueness of classical solutions is applicable to the equation u′′ − up = 0, p ∈ (1, ∞) , u (0) = u0, u′(0) = u1. (1)
Our study is motivated by the research on Chinese calligraphy. Neglect-ing the friction force of the paper on which a calligrapher creates his work through a handwritings brush (in Chinese, maue bie) with mass m (t) at time t, the displacement u (t) of the brush on reispaper (rice paper) at time t is governed by the Newtons’ second law of motion with the force F (t)
m (t) u′′
(t) = F (t) . (0.1.1)
Normally, the force F (t) depends on the displacement u (t) [4]1, that is F (t) = F (u (t)) . Experimentally, the change rate of the force is proportional to the change rate of displacement [4], that is, there is a real p so that
dF (t) dt F (t) = p du (t) dt u (t) .
By some calculation we find the form of the force F (u (t)) = cu (t)pfor some constant c.
Note that in the normal cases, and particularly for the beginner, the mass of their handwriting brushes vary with time due to the strength of hand and the intake of ink. For simplicity, we may assume that the mass depends upon the time periodically, in piecewise time interval; in other words,
m (t) = m1− k1t, 0 ≤ t ≤ t1, m2− k2t, t1 < t ≤ t2, .. . mn− knt, tn−1< t ≤ tn,
where kiand miare positive constants depend on the authors writingsusages.
To some calligraphers the mass of their brushes play no roll, and thus the mass of that brushes are all the same, in another words, m (t) = m for some constant m, therefore, the equation (0.1.1) becomes
u′′
(t) = c mu (t)
p.
(0.1.2)
If we set v (t) = (m/c)1/(p−1)u (t), then the equation (0.1.2) becomes v′′
(t) = v (t)p, in the form of (0.1) . Thus, the model of problem (0.1) describes a calligrapher with force up creating his works in real action. The initial
values u0 and u1 are non-negative. For p > 1, the null solution u (t) ≡ 0,
u0 = 0 = u1, corresponds to routine, uninspired works. When one is in an
outburst of enthusiasm for the writing, then in a short time there were some burned-curled-like curve would be created; in other words, for Eu(0) < 0 or
E (0) > 0 and u1 > 0, there exists a finite number T∗ such that u (t) −1
→ 0 as t → T∗
, c.f. Theorem 3 and 4.
From the observations, when the characteristic p of the calligrapher is smaller than 1, then their works could be good controlled or in some sense “nachmacht” (duplicated); mathematically, u (t) ≤ k (t ± c)θ, θ > 0.
These above-mentioned phenomena will be analyzed in the present paper mathematically bases on the model of the form (0.1).
We discuss the problem (0.1) in two parts, p > 1 and p < 1.
Part A. p > 1.
Notation and Fundamental Lemmas. For a given solution u (t) of (0.1) we set Eu(0) = u21− 2 p + 1u p+1 0 , Ju(t) = u (t) −p−1 2 .
Definition. A function g :R→R is said to have a blow-up rate q if g
exists only in finite time, that is, there is a finite number T∗
such that the following is valid lim t→T∗g (t) −1 = 0 (0.2)
and that there exists a non-zero β ∈R with
lim
t→T∗(T ∗
− t)qg (t) = β, (0.3)
in this case β is called the blow-up constant of g.
Since the solutions for the equation (0.1) is unique, we can rewrite Ju(t) = J (t) and Eu(t) = E (t) . From some elementary calculations we
obtain the following Lemma 1.
Lemma 1. Suppose that u is the solution of (0.1), then we have E (t) = u′ (t)2− 2 p + 1u (t) p+1 = E (0) , (0.4) (p + 3) u′ (t)2 = (p + 1) E (0) +u2(t)′′, (0.5)
J′′ (t) = p 2− 1 4 E (0) J (t) p+3 p−1 (0.6) and J′ (t)2= J′ (0)2−(p − 1) 2 4 E (0) J (0) 2(p+1) p−1 +(p − 1) 2 4 E (0) J (t) 2(p+1) p−1 . (0.7)
The following Lemma is easy to prove so we omit the arguments. Lemma 2. If g (t) and h (t, r) are continuous with respect to their variables and the limit limt→TR0g(t)h (t, r) dr exists, then
lim t→T Z g(t) 0 h (t, r) dr = Z g(T ) 0 h (T, r) dr.
I. Estimates for the life-spans. To estimate the life-span of the solution of the equation (0.1), we separate this section into three parts, E (0) < 0, E (0) = 0 and E (0) > 0. Here the life-span T of u means that u is the solution of problem (0.1) and the existence interval of u is [0, T ) so that the problem (0.1) has the solution u ∈ ¯C2(0, T ) and u make sense only
in this interval [0, T ).
I.1. E (0) ≤ 0. In this subsection we deal with the case that E (0) < 0 and E (0) = 0, u0u1 > 0. The case that E (0) = 0 and u0u1 ≤ 0 will be
considered in section 3 and section 4. We have the following result.
Theorem 3. If T is the life-span of u and u is the positive solution of the problem (0.1) with E (0) < 0, then T is finite. Further, for u0u1 ≥ 0 we
have T ≤ T∗ 1 (u0, u1, p) = 2 p − 1 Z J(0) 0 dr q k1+ E (0) rk2 ; (1.1.1)
for u0u1 < 0, T ≤ T∗ 2 (u0, u1, p) (1.1.2) = 2 p − 1 Z k 0 dr q k1+ E (0) rk2 + Z k J(0) dr q k1+ E (0) rk2 , where k1:= 2 p + 1, k2 := 2p + 2 p − 1 and k := 2 p + 1 −1 E (0) 2p+2p−1 . Furthermore, ifE (0) = 0 and u0u1> 0, then
T ≤ T∗ 3 := 2 p − 1 u0 u1 . (1.1.3)
Proof. Under the condition, E (0) < 0, we know immediately that u20> 0; otherwise we get u2
0 = 0, that is, u0 = 0, then E (0) = u21 ≥ 0; and this
contradicts to E (0) < 0. In this situation we divide the proof of the Theorem into two cases, u0u1≥ 0 and u0u1< 0.
(i) u0u1 ≥ 0. By identity (0.5) we find that
2uu′ (t) ≥ 2u0u1− (p + 1) E (0) t, ∀t ≥ 0, u2(t) ≥ u20+ 2u0u1t − p + 1 2 E (0) t 2, ∀t ≥ 0. (1.1.4)
From identity (0.7) , u0u1 ≥ 0 and the fact J′(t) = −p − 1
2 u (t) −p−1 2 u′(t) < 0, it follows that J′ (t) = −p − 12 q k1+ E (0) J (t)k2 ≤ J′(0) , ∀t ≥ 0, (1.1.5) where k1= u −p−1 0 u21− E (0) u 2−p+12 0 = 2 p + 1 and J (t) ≤ u− p−1 2 0 −p − 12 u −p+1 2 0 u1t, ∀t ≥ 0.
Thus, there exists a finite number T∗
1 (u0, u1, p) ≤ p−12 uu01 such that
J (T∗
1 (u0, u1, p)) = 0 and u (t) → ∞ for t → T1∗(u0, u1, p) . This means that
the life-span T of u is finite, that is, T ≤ T∗
1 (u0, u1, p) . Now we estimate
this life-span T∗
1 (u0, u1, p) .
By identity (1.1.5) and the fact that J (T∗
1 (u0, u1, p)) = 0 we find that Z J(0) J(t) dr q k1+ E (0) rk2 = p − 1 2 t, ∀t ≥ 0, (1.1.6)
and hence we get the estimate (1.1.1).
(ii) u0u1 < 0. For brevity, we only prove existence of critical point
t0(u0, u1, p) of u, that is, u′(t0(u0, u1, p)) = 0 and compute it later in section
III. By inequality (1.1.4), u0u1 < 0 and the convexity of u2 we can find a
unique finite number t0(u0, u1, p) such that
u (t) u′ (t) < 0 for t ∈ (0, t0(u0, u1, p)) , uu′ (t0(u0, u1, p)) = 0, uu′ (t) > 0 for t > t0(u0, u1, p) , (1.1.7)
and u (t0(u0, u1, p))2> 0. If not, then u (t0(u0, u1, p)) = 0, thus
E (0) = E (t0(u0, u1, p)) = u′(t0(u0, u1, p))2 ≥ 0;
yet this is in contradiction with E (0) < 0. Thus we conclude that
u2(t) > 0, ∀t ≥ 0. Hence we get u′
(t0(u0, u1, p)) = 0,
and J (t0(u0, u1, p))k2 = 2 p + 1 −1 E (0).
After arguments similar to the step (i), there exists a T∗
2 (u0, u1, p)
such that the life-span T of u is bounded by T∗
2 (u0, u1, p) , that is, T ≤
T∗
2 (u0, u1, p). On the analogy of the above argumentation, using (1.1.7) and
(0.7) we get J′ (t) = −p − 12 q k1+ E (0) J (t)k2, ∀t ≥ t0(u0, u1, p) , J′ (t) = p − 1 2 q k1+ E (0) J (t)k2, ∀t ∈ [0, t0(u0, u1, p)] . (1.1.8) Therefore we have Z J(t0) J(t) dr q k1+ E (0) rk2 = p − 1 2 (t − t0) , ∀t ≥ t0, Z J(t0) J(0) dr q k1+ E (0) rk2 = p − 1 2 t0. (1.1.9)
where t0 = t0(u0, u1, p) . Utilizing (1.1.9) and the fact that J (t0(u0, u1, p))k2
= p+12 −1
E(0) and J (T ∗
2 (u0, u1, p)) = 0 we obtain the estimate
T∗ 2 (u0, u1, p) = t0(u0, u1, p) + 2 p − 1 Z k 0 dr q k1+ E (0) rk2 . (1.1.10)
This estimate (1.1.10) is equivalent to (1.1.2).
(iii) E (0) = 0. Now we prove (1.1.3) . By identity (0.6) in Lemma 1 and E (0) = 0 we get J′′
(t) = 0 ∀t ≥ 0. From the positiveness of u0u1, it
follows that J′ (0) < 0 and J (t) = u− p−1 2 0 −p − 12 u −p+1 2 0 u1t, ∀t ≥ 0.
Thus we conclude that u (t) = u0 1 −p − 12 uu1 0 t − 2 p−1 , ∀t ≥ 0. (1.1.11)
Therefore the estimate (1.1.3) follows.
I.2. E (0) > 0, u0 ≥0.
In this subsection we consider two cases E (0) > 0, u0 > 0 and E (0) > 0,
u0 = 0, u1> 0.
We have the following blow-up result.
Theorem 4. Suppose that (i) u0 > 0 or
(ii) u0 = 0 and u1 > 0.
Then the life-span T of the positive solution u of the problem (0.1) with E (0) > 0 is finite, that is, u is only a local solution of (0.1).
Further, in case of(i) we have the estimates
T ≤ T∗ 4 (u0, u1, p) = 2 p − 1 Z J(0) 0 dr q k1+ E (0) rk2 , u1≥ 0; (1.2.1)
in the case of (ii)
T ≤ T∗ 5 (u0, u1, p) = 2 p − 1 Z ∞ 0 dr q k1+ E (0) rk2 . (1.2.2)
Proof. i) u0> 0. By identity (0.6) in Lemma 1 we obtain k3J′′(t) = (k3J (t))q, k3J (0) = k3u −p−1 2 0 , k3J′(0) = 1 − p 2 k3u −p+1 2 0 u1, (1.2.3) where k3 := p2−1 4 E (0) p−14 and q := p+3p−1. Now we set ˜ E (t) := k32J′ (t)2− 2 q + 1(k3J (t)) q+1,
after some calculations we see that ˜E (t) is a constant and ˜ E (t) = ˜E (0) = (p − 1) 2 4 k 2 3u −p−1 0 u21− E (0) (1.2.4)
From the condition that u0 > 0 and the definition of E (0) it follows
that 0 < ˜E (t) = (p − 1) 2 2 (p + 1)k 2 3u2(t) −p+3 2 u (t)p+3= (p − 1) 2 2 (p + 1)k 2 3, thus u (t)p+1> 0, ∀t ≥ 0. (1.2.5)
By identity (0.5) in Lemma 1 we find that u (t) u′ (t) = u0u1+ E (0) t + p + 3 p + 1 Z t 0 u (r) p+1 dr, ∀t ≥ 0 (1.2.6) and so u (t) u′ (t) ≥ u0u1+ E (0) t, ∀t ≥ 0. (1.2.7)
of Theorem 4 we get the conclusions (1.2.1) in Theorem 5.
Now let us show u0u1 ≥ 0. For u0u1 < 0, from (1.2.7) it follows that
u (t) u′
(t) ≥ 0 for large t. Suppose that ¯t0 is the first number such that
u (t) u′
(t) = 0. Using identity (0.5) in Lemma 1 we get (1.2.6.1) u (t) u′ (t) = E (0) (t − ¯t0) + p + 3 p + 1 Z t ¯ t0 u (r)p+1dr ≥ 0, ∀t ≥ ¯t0.
Hence we find that
uu′ (t) < 0 f or t ∈ (0, ¯t0) , uu′ (¯t0) = 0, uu′ (t) > 0 f or t > ¯t0, (1.2.8)
and u (¯t0) > 0; if not, then u (¯t0) = 0, this is in contradiction with (1.2.5) .
Hence we get
u′
(¯t0) = 0.
(1.2.9)
Therefore, by (1.2.5) we obtain that
(p + 1) E (0) = −2u (¯t0)p+1 < 0.
(1.2.10)
The identity (1.2.10) and the condition E (0) > 0 are in contradiction; therefore we get the assertion that u1 ≥ 0.
ii) By u0 = 0 and (1.2.6) we find
u (t) u′ (t) = E (0) t + p + 3 p + 1 Z t 0 u (r) p+1 dr, ∀t ≥ 0. (1.2.11) We claim that uu′
(t) > 0 for every t > 0. If not, then according to the positiveness of u1 there exists ˜t > 0 such that u ˜tu′ ˜t= 0. Let ˜T be the
first non-zero so that uT˜u′˜
again we get 0 = uu′˜ T= E (0) ˜T +p + 3 p + 1 Z T˜ 0 u (r) p+1dr.
This is therefore in contradiction with E (0) > 0; hence u (t) u′
(t) > 0 ∀t > 0 and J′
(t) < 0 ∀t > 0. Using (0.6) in Lemma 1 for each ˇt > 0 we conclude that J′ (t) = − s J′ ˇt2 −(p−1) 2 4 E (0) J ˇt2p+2 p−1 −J (t) 2p+2 p−1 , ∀t ≥ ˇt (1.2.12) and simultaneously lim ˇ t→0J ′ ˇ t2 −(p − 1) 2 4 u 2 1J ˇt 2p+2 p−1 = (p − 1) 2 2 (p + 1), thus by (1.2.12) , the estimate (1.2.2) follows.
I.3. Some properties concerning T∗
1 (u0, u1, p). In principle,
T∗
1 (u0, u1, p) depends on three variables u0, u1 and p. Set ck,p := (p+1)u 2 1 2up+10 , then T∗ 1 (u0, u1, p) = √ 2p + 2 p − 1 u −p−1 2 0 (1 − ck,p) −p−1 2p+2 Z (1−ck,p) p−1 2p+2 0 dr q 1 − r2p+2p−1 . It is evident that lim p→∞T ∗ 1 (u0, u1, p) = 0, lim p→∞T ∗ 1 (u0, u1, p) = ∞.
For convenience, we consider the case u1 = 0,
T∗ 1 (u0, 0, p) = √ π √2p + 2u−p−12 0 Γ2p+2p−1 Γp+1p .
Using Maple we get the graphs of T∗ 1 (u0, 0, p) below: Figure 1. Graph of T∗ 1 (u0, 0, p), u0∈ (0, 1), p ∈ [1, 5]. Figure 2. Graph of T∗ 1 (u0, 0, p), u0 ∈ [1, 50], p ∈ [1, 50].
Figure 3. Graphs of T∗
1 (u0, 0, p), u0≤ 1.
Figure 4. Graph of T∗
1 (u0, 0, p), u0 > 1.
The above pictures show the properties of T∗
1 (u0, 0, p):
(1) there exists a constant u∗
0 such that T ∗
in p for u0 ∈ [u∗0, 1);
(2) there is a p0 such that T1∗(u0, 0, p) is decreasing in (1, p0) and increasing
in (p0, ∞) provided u0∈ [0, u∗0);
(3) T∗
1 (u0, 0, p) is differentiable in its variables and
(4) for u0 > 1 the life-span T1∗(u0, 0, p) is decreasing in p.
We now show the validity of statements (3) and (4) using the mono-tonicity of T∗
1 (1, 0, p) for u0 6= 0. To prove (1) and (2) we must establish the
existence of u∗
0 with ∂p∂ T ∗
1 (u0, 0, p) ≤ 0 for 1 > u0 ≥ u∗0, that is,
0 ≤ p − 1p + 1(p + 3) Z 1 0 1 − r2p+1p−1 −1/2 dr +4 Z 1 0 1 − r2p+1p−1 −3/2 r2p−1p+1ln r dr + (p − 1)2(ln u0) Z 1 0 1 − r2p+1p−1 −1/2 dr, thus the existence of u∗
0 can be obtained provided
p − 1 p + 1(p + 3) r2p+1p−1 − 1 − 4 ln r > 0, ∀r > 1. After some calculations it is easy to get the above assertion.
To grasp the property of the life-span T∗
1 (u0, u1, p) is very difficult, but
for fixed initial data we want to know how the life-span varies with p, so now we consider the life-span T∗
1 (0.6, 0.2, p) and list the following tables as
below. p T∗ 1 (0.6, 0.2, p) 1.001 2001. 5 1.004 501. 42 1.008 251. 42 1.012 168. 08 p T∗ 1 (0.6, 0.2, p) 2 3. 4135 2.5 2. 7698 3 2. 4659 3.6497 2. 2644
After some computations we get T∗ 1 (u0, u1, p) = √ 2p + 2 p − 1 up+10 −p + 1 2 u 2 1 −p−1 2p+2 Z (1−p+12 u−p−10 u21) p−1 2p+2 0 dr q 1 − r2p+2p−1 .
By the experience in studying the life span T∗
1 (u0, 0, p) , we consider the
properties of the life-span T∗
1 (u0, u1, p) with u0u1≥ 0 in three cases:
Case 1: 0 < up+10 − (p + 1) u2
1/2 < 1. In this situation we find that
(i) for fixed u1,
(5) there exists a constant u∗
0 depending on u1 such that T1∗(u0, u1, p) is
monotone decreasing in p for u0≥ u∗0,
(6) there is a p0 so that T1∗(u0, u1, p) decreases in (1, p0) and increases in
(p0, ∞) provided u0∈ [0, u∗0);
(ii) for fixed u0, the life-span T1∗(u0, u1, p) decreases in u21.
Case2: up+10 −(p+1)u21/2 > 1. The life-span T∗
1(u0, u1, p) decreases in p. Case 3: up+10 − (p + 1) u2 1/2 = 1. On the surface n (u0, u1, p) ∈R3 u p+1 0 − (p + 1) u21/2 = 1, p > 1 o we find that T∗ 1 (u0, u1, p) = T1∗(u0, p) = √ 2p + 2 p − 1 Z u−(p−1)/20 0 1 p 1 − r2(p+1)/(p−1)dr and T∗
1 (u0, p) is monotone decreasing in u0 and in p.
II. Blow-up rate and blow-up constant. In this section we study the blow-up rate and blow-up constant for u2, u2′
and u2′′
under the conditions in section 1. We have the following results.
Theorem 5: If u is the positive solution of the problem (0.1) with one of the following properties that
(i) E (0) < 0 or (ii) E (0) = 0, u0u1 > 0 or (iii) E (0) > 0, u0> 0 or (iv) E (0) > 0, u0= 0, u1 > 0
Then the blow-up rate of u is 2/ (p − 1) , and the blow-up constant of u is p−1q
2 (p − 1)−2
(p + 1), that is, for m ∈ {1, 2, 3, 4, 5, 6} lim t→T∗ m(u0,u1,p) (T∗ m(u0, u1, p)−t) 2 p−1u (t) = 2 1 p−1 (p + 1) 1 p−1(p−1)− 2 p−1 . (2.1.1)
The blow-up rate of u′
is (p + 1) / (p − 1), and the blow-up constant of u′
is 2p−1p (p + 1) 1
p−1(p − 1)− p+1
p−1, that is, for m ∈ {1, 2, 3, 4, 5, 6} .
lim t→T∗ m(u0,u1,p) u′ (t) (T∗ m(u0, u1, p) − t) p+1 p−1 (2.1.2) = 2p−1p (p + 1) 1 p−1(p − 1)− p+1 p−1.
The blow-up rate of u′′
is 2p/ (p − 1), and the blow-up constant of u′′
is 2p−1p (p + 1)
p
p−1(p − 1)− 2p
p−1, that is, for m ∈ {1, 2, 3, 4, 5, 6}
lim t→T∗ m(u0,u1,p) u′′ (t) (T∗ m(u0, u1, p)−t) 2p p−1 (2.1.3) = 2p−1p (p + 1) p p−1(p−1)− 2p p−1.
(1.1.6) we get Z J(t) 0 1 T∗ 1 (u0, u1, p) − t dr q k1+ E (0) rk2 = p − 1 2 , ∀t ≥ 0. (2.1.4)
By Lemma 4 and (2.1.4) we obtain limt→T∗ 1(u0,u1,p) 1 √ k1 J (t) T∗ 1 (u0, u1, p) − t = p − 1 2 . (2.1.5)
This identity (2.1.5) is equivalent to (2.1.1) for m = 1. For E (0) < 0, u0u1 < 0 using (1.1.9) we have also
Z J(t) 0 dr q k1+E (0) rk2 = p−1 2 (T ∗ 2(u0, u1, p)−t) , ∀t≥t0(u0, u1, p). (2.1.6)
From the Lemma 4 and (2.1.6) , the estimate (2.1.1) for m = 2 follows. Utilizing the identities (1.1.5) and (1.1.8) we find
lim t→T∗ m(u0,u1,p) J′ (t) = −√p − 1 2p + 2, m = 1, 2. (2.1.7)
Therefore, by (2.1.7) we have for m = 1, 2
lim t→T∗ m(u0,u1,p) u2′(t) (T∗ m(u0, u1, p) − t) p+3 p−1 (2.1.8) = 2p−12p (p + 1) 2 p−1(p − 1)− p+3 p−1,
and thus, for m = 1, 2
lim t→T∗ m(u0,u1,p) u′ (t)2(T∗ m(u0, u1, p) − t) 2p+2 p−1 (2.1.9) = 2p−12p (p + 1) 2 p−1(p − 1)− 2p+2 p−1 .
Through (0.5) and (2.1.9) for m = 1, 2, we obtain the estimate lim t→Tm∗(u0,u1,p) u2′′(t) (T∗ m(u0, u1, p) − t) 2p+2 p−1 = (p + 3) lim t→T∗ m u′ (t)2(T∗ m(u0, u1, p) − t) 2p+2 p−1 , (2.1.10) lim t→T∗ m(u0,u1,p) 2u (t) u′′ (t) (T∗ m(u0, u1, p) − t) 2p+2 p−1 = (p + 1) lim t→T∗ m u′ (t)2(T∗ m(u0, u1, p) − t) 2p+2 p−1 = 2p−12p (p + 1) p+1 p−1(p − 1)− 2p+2 p−1 and lim t→T∗ m(u0,u1,p) u′′ (t) (T∗ m(u0, u1, p) − t) 2p p−1 = 2 p p−1(p + 1) p p−1(p − 1)− 2p p−1
Thus the estimate (2.1.3) for m = 1, 2 is proved.
ii) For E (0) = 0, u0u1 > 0, for m = 3, using identity (1.1.11) we get
u2(t) = u2 p+3 p−1 0 (p − 12 u0u1) − 4 p−1(T∗ m(u0, u1, p) − t) − 4 p−1, ∀t ≥ 0. (2.1.11)
Therefore the estimates (2.1.1) , (2.1.2) and (2.1.3) for m = 3 follow from (2.1.11).
iii) The estimates (2.1.1) , (2.1.2) and (2.1.3) for m = 4, 5 are similar to the above arguments (i) in the proof of this Theorem.
Now we consider the property of the blow-up constants K1, K2 and K3.
We have K1(p) = 2 1 p−1(p + 1)p−11 (p − 1)−p−12 , K2(p) = 2 p p−1(p + 1) 1 p−1 (p − 1)− p+1 p−1 , K3(p) = 2 p p−1(p + 1) p p−1 (p − 1)− 2p p−1 .
Figure 5. Graph of K1(p), K2(p), K3(p)
Figure 6. Graph of K1(p), K2(p), K3(p)
We see that the graphs, Ki(p), i = 1, 2, 3 are all decreasing in p ∈ (1, p1) ;
and Ki(p) tends to zero for i = 2, 3 and K1(p) tends to 1, as p tends to
infinity. The monotonicity of these functions can be obtained after showing the following inequalities:
d dpK1(p) = (2p + 2) 1 p−1(p − 1)− 2 p−1−2 ln(p−1) 2 2p + 2− p + 3 p + 1 ! ≤ 0, p ∈ (1, p1) where p1∼ 9.2203,
14 -0.1 6 8 10 12 4 0 -0.02 -0.04 -0.06 -0.08 Figure 7. Graph of dpdK1(p). 14 -40 2 4 6 8 10 12 0 -10 -20 -30 Figure 8. Graph of ln(p−1)2p+22 −p+3p+1. p + ln (2p + 2) + 2 p + 1 ≥ 2 ln (p − 1) , ∀p > 1. The above inequality is easy to prove, we omit the arguments.
III. Uniqueness on p and extension. In practical the characteristic index p (t) depends on the characteristic (at time t) of the calligrapher
him-self only, in other words, when two “Werke” are similar to each other, then the correspondent characteristic p (t) must very close, in mathematics, the fact can be easily solved to the scalar constant p (t) = p, we write it below.
Theorem 6. Suppose that u and v are the positive solutions of the following equations respectively
u′′
(t) = u (t)p, v′′
(t) = v (t)q (3.1)
with u (t) 6= 0 6= v (t) for each t ≥ 0. If they have the same rate of displace-ment, that is,
u′
(t) /u (t) = v′
(t) /v (t) , (3.2)
then they posses the same characteristic, this means, p = q. Proof. According to the condition (3.2), we have
u (t)p+1− u′ (t)2 u (t)2 = v (t)q+1− v′ (t)2 v (t)2 . Using (3.2) again, then
u (t)p−1= v (t)q−1.
This together with (3.2) we obtain the assertion. For E (0) = 0, u0u1 < 0, it is easy to see that
u (t) = u p+3 p−1 0 u20− p − 1 2 u0u1t − 2 p−1 , ∀t ∈ (0, T ) . Hence we find the limit limt→∞u (t) = 0 and
lim t→∞t 2 p−1u (t) = u p+3 p−1 0 p − 1 −2 u0u1 − 2 p−1 .
The following Theorem is a direct application of Theorem 4, Theorem 6 and we omit the proof.
Theorem 7. If u ∈ P C2(R+), that is, u ∈C2(S∞
i=0(Ti, Ti+1) ∪ (T∞, ∞))
where T0 = 0, Ti+1 ≥ Ti and T∞ = limi→∞Ti, is a piecewise solution of
the problem of (0.1) with E (t) < 0 for the continuous points of E. Then for T∞= ∞, the discontinuous points of u can be got at the blow-up points
¯ T∗
m(u0, u1, p) , m ∈N of u2(t) and ¯Tm∗ (u0, u1, p) are given by
¯ T∗ 1 (u0, u1, p) := 2T∗ 1 (u0, u1, p) if u0u1 ≥ 0 and uu′(T1∗+(u0, u1, p)) ≥0, (T∗ 1 + T2∗)(u0, u1, p) if u0u1< 0 and uu′(T1∗+(u0, u1, p)) ≥0, 2T∗ 2 (u0, u1, p) if u0u1< 0 and uu′(T ∗+ 1 (u0, u1, p)) ≥0 (3.3) and ¯ T∗ m+1(u0, u1, p) := ¯ T∗ m+ T1∗ (u0, u1, p) if uu′ T¯m∗+(u0, u1, p)≥ 0, ¯ T∗ m+ T ∗ 2 (u0, u1, p) if uu′ T¯m∗+(u0, u1, p)< 0, (3.4) where uu′ ¯ T∗+ m (u0, u1, p):= limt→T∗+ m+7 u2(t)−u(T¯∗ m(u0,u1,p))2 t− ¯Tm∗(u0,u1,p) . Further we have the blow-up rate at ¯T∗
m(u0, u1, p) of u2 is 4/ (p − 1) , and
the blow-up constant of u2 is p−1
q
4 (p − 1)−4
(p + 1)2, that is, for m ∈N
limt→T∗ m T¯ ∗ m(u0, u1, p) − t 4 p−1u2(t) = 2p−12 (p + 1) 2 p−1 (p − 1)− 4 p−1 . (3.5)
The blow-up rate of u2′
at ¯T∗
m(u0, u1, p) is (p + 3) / (p − 1), and the
blow-up constant of u2′
is2p−12p (p + 1) 2
p−1(p − 1)− p+3
lim t→T∗ m(u0,u1,p) ¯ T∗ m(u0, u1, p) − t p+3 p−1u2 ′ (t) (3.6) = 2p−12p (p + 1)p−12 (p − 1)− p+3 p−1.
The blow-up rate of u2′′
at ¯T∗
m(u0, u1, p) is (2p + 2) / (p − 1), and the
blow-up constant of u2′′
is2p−12p (p + 1)p−18 (p − 1)− 2p+8
p−1 (p + 3), that is, for m ∈
N lim t→T∗ m(u0,u1,p) u2 ′′ (t) (T∗ m(u0, u1, p) − t) 2p+2 p−1 (3.7) = 2 p − 1 2p p−1 (p + 3) p + 1 p − 1 2 p−1 .
Part B. Positive solution for p < 1. Before the study of the prop-erties of solutions for the differential equation (0.1) we collect some results on the situation that Eu(0) = 0.
(1) For u0> 0 and u1 > 0, we have
u (t) = u 1−p 2 0 +1 − p2 s 2 p + 1t !1−p2 and tp−12 u (t) → 1 − p 2 s 2 p + 1 !1−p2 as t → ∞.
(2) For u0> 0 and u1 < 0, the solutions of (0.1) can be given as
uc(t) = u1−p2 0 +1−p2 q 2 p+1t 1−p2 t ∈ [0, T0] 0 t ∈ [T0, T0+ c] (1−p)2 2p+2 1−p1 (t − T0− c) 2 1−p t ≥ T0+ c
where c is any positive real number and T0 = q p+1 2 u 1−p 0 , and also tp−12 u (t) → 1 − p 2 s 2 p + 1 !1−p2 as t → ∞.
IV. Eu(0) > 0. In this section we discuss the case Eu(0) > 0 and we
have the following result concerning the zero point and asymptotic behavior at infinity of the solutions for the equation (0.1) :
Theorem 8. Suppose that T∗
is the life-span of u which is a positive solution of problem (0.1) with Eu(0) > 0 and u0 > 0. Then for
(1) u1 < 0, there exists a constant Z0 so that T∗ ≤ Z0 and limt→Z0u (t) = 0, limt→Z0u ′ (t) = −p Eu(0) and limt→Z0u ′′′ (t)−1 = 0. Moreover, Z0 = Z u0 0 dr q Eu(0) + p+12 rp+1 , (4.1) lim t→Z− 0 u′′′ (t) (t − Z0)1−p= pEu(0) p 2; (4.2) (2) u1> 0, lim t→∞u (t) t − 2 1−p = 1 − p 2 s 2 p + 1 !1−p2 . (4.3)
Proof. (1) For u1 < 0, after some calculations we obtain
u′ (t) = − s Eu(0) + 2 p + 1u (t) p+1 ≤ − s 2 p + 1u (t) p+1 , ∀t ∈ [0, T∗ ) (4.4) and u (t) ≤ u1−p2 0 −1 − p2 t 2 1−p , ∀t ∈ [0, T∗ ) ;
thus there exists a constant Z0 so that T∗≤ Z0 and limt→Z0u (t) = 0. By (4.4) we conclude that limt→Z−
0 u ′ (t) = −p Eu(0) and t = Z u0 u(t) dr q Eu(0) + p+12 rp+1 , ∀t ∈ [0, T∗ ) , Z0 = lim t→Z0 Z u0 u(t) dr q Eu(0) + p+12 rp+1 = Z u0 0 dr q Eu(0) + p+12 rp+1 and lim t→Z− 0 u′′′ (t) (t − Z0)1−p = p lim t→Z− 0 u (t) t − Z0 p−1 u′ (t) = pEu(0) p 2 .
Therefore (4.1) and (4.2) are proved. (2) For u1> 0 we have u′ (t) = s Eu(0) + 2 p + 1u (t) p+1 ≥ s 2 p + 1u (t) p+1, ∀t ≥ 0, u (t)1−p2 ≥ u 1−p 2 0 +1 − p2 s 2 p + 1t, ∀t ≥ 0. (4.5)
On the other hand,
u′ (t) ≤ s 2 p + 1 u (t) + p + 1 2 Eu(0) p+11 ! p+1 2 , ∀t ≥ 0, u (t) + p + 1 2 Eu(0) p+11 ! 1−p 2 (4.6) ≤ u0+ p + 1 2 Eu(0) p+11 ! 1−p 2 +1 − p 2 s 2 p + 1t ∀t ≥ 0. From (4.5) and (4.6), the estimate (4.3) follows.
V. Eu(0) < 0. In this section we discuss the case Eu(0) < 0. Similar
to the above arguments proving Theorem 8 we have the following result on critical point and asymptotic behavior at infinity of the solutions for the equation (0.1) :
Theorem 9. Suppose that u is a positive solution of problem (0.1) with Eu(0) < 0 and u0> 0. Then lim t→∞u (t) t − 2 1−p = 1 − p 2 s 2 p + 1 !1−p2 . (5.1)
Moreover, for u1 < 0, there exists a constant Z1 so that limt→Z1u ′ (t) = 0 and Z1 = p+1 r p + 1 2 (−Eu(0)) 1−p 2p+2 Z (p+1 −2Eu(0)) −1 p+1u0 1 dr √ rp+1− 1. (5.2)
Remark. We do not know whether the solutions under the circum-stance in Theorem 9 is analytic or not.
Through Theorems 3 through 7 may be summarized for p > 1, in the following tables E (0) E (0) < 0 E (0) = 0 T (i) u0u1≥ 0, T ≤ T ∗ 1 (u0, u1, p) (ii) u0u1< 0, T ≤ T2∗(u0, u1, p) (i) u0u1 > 0, T ≤ T3∗ (ii) u0u1< 0, T = ∞ (iii) u0u1= 0, T = ∞, u ≡ 0. R1, K1 4 p − 1, K1 (p) 4 p − 1, K1 (p) R2, K2 p + 3 p − 1, K2 (p) p + 3 p − 1, K2 (p) R3, K3 2p + 2 p − 1 , K3 (p) 2p + 2 p − 1 , K3 (p)
E (0) > 0 E (0) > 0ˆ E (0) = 0, uˆ 1> 0 T T ≤ T∗ 4 (u0, u1, p) T ≤ T5∗(u0, u1, p) R1, K1 4 p − 1, K1 (p) 4 p − 1, K1 (p) R2, K2 p + 3 p − 1, K2 (p) p + 3 p − 1, K2 (p) R3, K3 2p + 2 p − 1 , K3 (p) 2p + 2 p − 1 , K3 (p)
Where T := Lif e − span of u, E (0) = Energy, R1= blow − up rate of
a, K1 = blow − up constant of a; R2 = blow − up rate of a′, K2 = blow − up
constant of a′
; R3 = blow − up rate of a′′, K3 = blow − up constant of a′′;
ˆ
E (0) := u20u21− 4u20E (0) .
Acknowledgment.
I want to thanks to professor Tsai Long-Yi and professor Chen Tain-Chian for their continuously encouragement and their discussions to this work for the writing, to NSC and Grand Hall for their financial support and to the referee for his interesting and helpful comments to this work.
References
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2. M.-R. Li, Estimations for the life-span of solutions for semilinear wave equations, Proceedings of the Workshop on Differential Equations V. Jan. 10-11, 1997, National Tsing-Hua Uni. Hsinchu, Taiwan.
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Department of Mathematical Sciences, National Chengchi University, Taipei 116, Taiwan, R.O.C.
